Question

A body is executing Simple Harmonic Motion. Find the average potential energy of the body.
(A) $\dfrac{1}{2}k{a^2}$
(B) $\dfrac{1}{4}k{a^2}$
(C) $k{a^2}$
(D) Zero

Answer 1

Hint The average potential energy or average kinetic energy of a body is equal to half of the total energy in simple harmonic motion. It is given by,
$E = \dfrac{1}{2}m{\omega ^2}{A^2}$
where, $A$ is amplitude and Total energy is equal to the sum of both kinetic energy and potential energy.

Complete step by step answer:
The motion in which the restoring force is directly proportional to the displacement of the body from its mean position. The direction of restoring force is always towards the mean position. The particle which executes simple harmonic motion has acceleration, $a\left( t \right) = - {\omega ^2}x\left( t \right)$
Here, $\omega $ is the angular velocity of the particle.
The total energy of the particle in simple harmonic motion is calculated by the formula –
$E = \dfrac{1}{2}m{\omega ^2}{A^2}$
where, $A$ is amplitude and remains conserved, and
$\omega $ is the angular velocity of the particle.
Total energy of the particle in simple harmonic motion is equal to the sum of kinetic energy and potential energy of the particle. So, -
$E = K + U$
Therefore, the average potential energy or kinetic energy can be calculated by –
\[
  {K_{avg}} = {U_{avg}} = \dfrac{E}{2} \\
  {K_{avg}} = {U_{avg}} = \dfrac{{\dfrac{1}{2}m{\omega ^2}{A^2} + \dfrac{1}{2}m{\omega ^2}{A^2}}}{2} \\
  {K_{avg}} = {U_{avg}} = \dfrac{1}{4}m{\omega ^2}{A^2} \\
 \]
So, we got the expression for average potential energy for the particle in simple harmonic motion which can be expressed as –
\[{U_{avg}} = \dfrac{1}{4}m{\omega ^2}{A^2} \cdots \left( 1 \right)\]
We know that, in the above $m{\omega ^2}$ is the constant value. So, let constant be $k$.
$\therefore m{\omega ^2} = k$
Putting this value of $m{\omega ^2}$ in equation $\left( 1 \right)$, we get –
\[{U_{avg}} = \dfrac{1}{4}k{A^2}\]
Hence, the average potential energy of particle in simple harmonic motion can be expressed as –
\[{U_{avg}} = \dfrac{{k{A^2}}}{4}\]

Therefore, the correct option is (B).

Note The energy which is stored in the particle or an object is called Potential energy. The S.I unit of potential energy is Joules.
The average kinetic energy of the particle in simple harmonic motion can also be known by –
${K_{avg}} = \dfrac{1}{T}\int\limits_0^T {Kdt} $
Similarly, the average potential energy of the particle in simple harmonic motion can be known by –
${U_{avg}} = \dfrac{1}{T}\int\limits_0^T {Kdt} $